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Answer to Question #176462 in General Chemistry for FlorenceMusonda

Question #176462

1liter of 0.400M NH3 solution also contains 12.78g of NH4Cl .How Much will be the pH of this solution change if 0.142mol of gaseous HCl is bubbled into it? Kb for NH3 is 1.8*10^-5

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Expert's answer
2021-03-29T06:09:18-0400

Number of moles for NH4Cl = 12.78/53.491 g/mol = 0.2376

Ka = 1×10^ -14/ 1.8 ×10 ^ -5 = 5.56×10^-10

0.4 × 1000 / 17.031 = 23.53 moles

Hydrolysis of NHCl -net ionic equation

Nh4++H2O -> NH3 + H3O

Using Ka equation

23 NH3 = 0.142 (H3O+) and NH4 + = 0.2375 M

= 0.142/23 = 0.0068

= 0.0068 (H3O +) ^2 = (5.56 × 10^-10) × 0.2376

0.0068 (H3O +)^2= (1.3211 × 10^ -10)/ 0.0068

(H3O +) ^ 2= 1.96 × 10^-8

H3O+ = 1.4 × 10^-4

PH = - log( H30+)

= - log (1.4 × 10 ^-4)

PH = 3.85

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